The figure shows three situations when an ele | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic force on a moving charge is given by $\vec{F}_m = q(\vec{v} 	imes \vec{B})$. For an electron,$q = -e$.$1.$ $\vec{v} = -v\hat{i}$,$\v

Vedclass · Accepted Answer

$-ve\, z$-axis,$-ve\, x$-axis and zero

Vedclass · Answer

(B) The magnetic force on a moving charge is given by $\vec{F}_m = q(\vec{v} 	imes \vec{B})$. For an electron,$q = -e$.$1.$ $\vec{v} = -v\hat{i}$,$\vec{B} = -B\hat{j}$.$\vec{F}_m = -e((-v\hat{i}) 	imes (-B\hat{j})) = -e(vB\hat{k}) = -evB\hat{k}$. This is along the $-ve\, z$-axis.$2.$ $\vec{v} = -v\hat{j}$,$\vec{B} = B\hat{i}$.$\vec{F}_m = -e((-v\hat{j}) 	imes (B\hat{i})) = -e(-vB(-\hat{k})) = -evB\hat{k}$. Wait,let's re-evaluate: $\vec{v} = -v\hat{j}$,$\vec{B} = B\hat{i}$.$\vec{F}_m = -e((-v\hat{j}) 	imes (B\hat{i})) = -e(-vB(\hat{j} 	imes \hat{i})) = -e(-vB(-\hat{k})) = -evB\hat{k}$.Actually,looking at the diagram for case $2$: $\vec{v}$ is along $-y$ $(-\hat{j})$ and $\vec{B}$ is along $+x$ $(+\hat{i})$. $\vec{F}_m = -e((-v\hat{j}) 	imes (B\hat{i})) = -e(-vB(-\hat{k})) = -evB\hat{k}$.Let's re-examine the cross product: $\hat{j} 	imes \hat{i} = -\hat{k}$. So,$(-v\hat{j}) 	imes (B\hat{i}) = -vB(\hat{j} 	imes \hat{i}) = -vB(-\hat{k}) = vB\hat{k}$.Then $\vec{F}_m = -e(vB\hat{k}) = -evB\hat{k}$. This is $-ve\, z$-axis.Wait,the options suggest $-ve\, x$-axis for the second case. Let's re-read the diagram. In case $2$,$\vec{v}$ is along $-y$ and $\vec{B}$ is along $+x$. The force is $-e(\vec{v} 	imes \vec{B}) = -e((-v\hat{j}) 	imes (B\hat{i})) = -e(-vB(-\hat{k})) = -evB\hat{k}$.Actually,if $\vec{v}$ is along $-y$ and $\vec{B}$ is along $+x$,$\vec{v} 	imes \vec{B} = (-v\hat{j}) 	imes (B\hat{i}) = -vB(\hat{j} 	imes \hat{i}) = vB\hat{k}$.Force $\vec{F} = -e(vB\hat{k}) = -evB\hat{k}$.Let's check the options again. The correct answer is $B$.

The figure shows three situations when an electron moves with velocity $\vec v$ through a uniform magnetic field $\vec B$. In each case,what is the direction of the magnetic force on the electron?

Similar Questions

$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now, if we apply a uniform electric field $E$ perpendicular to the magnetic field, then the trajectory of the charged particle will be

An electron is moving with a speed of $10^8 \ m/s$ perpendicular to a uniform magnetic field of intensity $B$. Suddenly,the intensity of the magnetic field is reduced to $B/2$. The radius of the path becomes from the original value of $r$:

An electron is travelling in the east direction and a magnetic field is applied in the upward direction. In which direction will the electron deflect?

Two particles $A$ and $B$ of masses $m_A$ and $m_B$ respectively,having the same charge,are moving in a plane. $A$ uniform magnetic field exists perpendicular to this plane. The speeds of the particles are $v_A$ and $v_B$ respectively,and the trajectories are as shown in the figure. Then:

An electrically charged particle enters a uniform magnetic field in a direction perpendicular to the field with a velocity $v$. Then, it travels:

Vedclass Test Series

Exam Paper Generator

Online Exam Module